Shaskolskaya & Eltsin Problem 205

Physics · thermodynamics · heat transfer

Problem

Since there are no losses of heat through the sides of the cylinders, the amount of heat which flows through any section of our system in unit time will be the same. If the difference in temperature between any two cross-sections of a homogeneous medium, distance l apart, is T₁ - T₂, the amount of heat which flows from the first to the second in unit time through any section between them will be expressed as Q = kS (T₁ - T₂)/l, where k is the coefficient of thermal conductivity and S is the sectional area. Let T₁ be the temperature of the upper end of the iron cylinder, T₂ be the temperature of the ends in contact, T₃ be the temperature of the lower end of the silver cylinder. Then the amount of heat which flows through any section of the iron cylinder equals Q_Fe = k_Fe S (T₁ - T₂)/l. Similarly for any section of the silver cylinder we shall have Q_Ag = k_Ag S (T₂ - T₃)/l. Comparing these expressions, we shall get: k_Fe(T₁ - T₂) = k_Ag(T₂ - T₃). Substituting the values of T₁ = 100°C, T₃ = 0°C and k_Ag = 11k_Fe in the above equation we find that the temperature of the ends in contact is T₂ = 100/12 = 8.33°C. 206. Birch wood burns faster than pine wood and therefore more heat is given off from it in unit time. HEAT AND CAPILLARY PHENOMENA 205

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